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$$\begin{matrix} CH_2 &-&CH&-&CH&-&CH&-&CH_2  \\|& &|&&|&&|&&|  \\ OH&&OH&&OH&&OH&&OH  \end{matrix}$$

The number of configurational isomers of the above-given compound is:


Solution

As we know, the number of optically active forms is $$=a=2^{n-1}-2^{\frac{n-1}{2}};  a=2^{3-1}-2^{\frac{3-1}{2}}=4-2=2$$, 
where n is the number of chiral centres, which is 3 here.
The number of meso forms, m is $$=2^{\frac{3-1}{2}}=2^{1}=2$$.
The total number of configurational isomers is, $$a + m = 2 + 2 = 4$$.

So, the correct answer is 4

Chemistry

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